- Determine the slope and equation for each individual line in every pair using two given points on each line.
- Set the equations of each pair of lines equal to each other to solve for the \(x\)-coordinate of the intersection point.
- Substitute the \(x\)-coordinate back into one of the original line equations to find the corresponding \(y\)-coordinate for each intersection.
- The identified intersection points are \((-4, -7)\), \((4, 7)\), \((-4, -5)\), and \((1, -1)\).
- Final answer: \(\boxed{(-4, -7), (4, 7), (-4, -5), (1, -1)}\)
Explanation
- Problem Analysis and Methodology
We need to determine the intersection points of four pairs of lines, where each line is defined by two points. The equation of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\) is found using the slope \(m = \frac{y_2 - y_1}{x_2 - x_1}\) and the point-slope form \(y - y_1 = m(x - x_1)\). The intersection point \((x, y)\) is the point that satisfies the equations of both lines simultaneously.
- Calculation of Intersection Points
- Pair 1:
Line 1: Passes through \((2, 5)\) and \((-3, -5)\). Slope \(m_1 = \frac{-5 - 5}{-3 - 2} = \frac{-10}{-5} = 2\). Equation: \(y - 5 = 2(x - 2) \Rightarrow y = 2x + 1\).
Line 2: Passes through \((3, 0)\) and \((0, -3)\). Slope \(m_2 = \frac{-3 - 0}{0 - 3} = \frac{-3}{-3} = 1\). Equation: \(y = x - 3\).
Intersection: \(2x + 1 = x - 3 \Rightarrow x = -4\). Substituting \(x = -4\) into \(y = x - 3\), we get \(y = -7\). Point: \((-4, -7)\).
- Pair 2:
Line 1: Passes through \((1, 1)\) and \((2, 3)\). Slope \(m_1 = \frac{3 - 1}{2 - 1} = 2\). Equation: \(y - 1 = 2(x - 1) \Rightarrow y = 2x - 1\).
Line 2: Passes through \((0, 3)\) and \((2, 5)\). Slope \(m_2 = \frac{5 - 3}{2 - 0} = 1\). Equation: \(y = x + 3\).
Intersection: \(2x - 1 = x + 3 \Rightarrow x = 4\). Substituting \(x = 4\) into \(y = x + 3\), we get \(y = 7\). Point: \((4, 7)\).
- Pair 3:
Line 1: Passes through \((1, 0)\) and \((0, -1)\). Slope \(m_1 = \frac{-1 - 0}{0 - 1} = 1\). Equation: \(y = x - 1\).
Line 2: Passes through \((0, 3)\) and \((-2, -1)\). Slope \(m_2 = \frac{-1 - 3}{-2 - 0} = \frac{-4}{-2} = 2\). Equation: \(y = 2x + 3\).
Intersection: \(x - 1 = 2x + 3 \Rightarrow -x = 4 \Rightarrow x = -4\). Substituting \(x = -4\) into \(y = x - 1\), we get \(y = -5\). Point: \((-4, -5)\).
- Pair 4:
Line 1: Passes through \((2, 0)\) and \((0, -2)\). Slope \(m_1 = \frac{-2 - 0}{0 - 2} = 1\). Equation: \(y = x - 2\).
Line 2: Passes through \((4, 5)\) and \((3, 3)\). Slope \(m_2 = \frac{3 - 5}{3 - 4} = 2\). Equation: \(y - 3 = 2(x - 3) \Rightarrow y = 2x - 3\).
Intersection: \(x - 2 = 2x - 3 \Rightarrow -x = -1 \Rightarrow x = 1\). Substituting \(x = 1\) into \(y = x - 2\), we get \(y = -1\). Point: \((1, -1)\).
- Conclusion
Matching the results to the provided list of intersection points:
- Pair 1: \((-4, -7)\)
- Pair 2: \((4, 7)\)
- Pair 3: \((-4, -5)\)
- Pair 4: \((1, -1)\)
The final intersection points are summarized above.
Examples
In urban planning, identifying the intersection point of two paths or roads is crucial for determining the location of a junction, a traffic light, or a landmark. By representing paths as linear equations based on GPS coordinates or map locations, planners use systems of equations to find where these paths cross, ensuring accurate placement of infrastructure and optimal traffic management.